IPMAT Indore 2023 · Quantitative Ability · Question 25

IPMAT Indore 2023 previous year question on Permutation and Combination, with the answer and a step-by-step solution.

Permutation and CombinationModern MathsMultiple choiceMedium
A rabbit is sitting at the base of a staircase which has 10 steps. It proceeds to the top of the staircase by climbing either one step at a time or two steps at a time. The number of ways it can reach the top is
  1. A
    34
  2. B
    89
  3. C
    144
  4. D
    55
Pick an option, then check.
Answer & solution

Answer: Option B

Step 1 — set up the recurrence. Let {W}_{n} be the number of ways to reach step n. The rabbit’s last move was either a 1-step (from step n-1) or a 2-step (from step n-2), and those cases are disjoint:

{W}_{n}={W}_{n-1}+{W}_{n-2}

Step 2 — fix the base cases.

{W}_{1}=1 \left( \text{one single step}\right)

{W}_{2}=2 \left( 1+1 \text{or}2 \right)

Step 3 — climb.

n

1

2

3

4

5

6

7

8

9

10

{W}_{n}

1

2

3

5

8

13

21

34

55

89

Step 4 — read off. {W}_{10}=89.

This is the Fibonacci sequence; in the standard indexing {W}_{n}={F}_{n+1}, so {W}_{10}={F}_{11}=89.

Answer: Option 2 — 89

Shortcut. Recognise Fibonacci on sight and count carefully from the right base. The sequence of answers for n=1,2,3,\ldots is 1,2,3,5,8,13,21,34,55,89 — so 10 steps gives 89.

Every wrong option is a Fibonacci number, which is the whole point: 55 is the answer for 9 steps, 144 for 11, and 34 for 8. Getting the indexing right is the entire question, so always anchor on the two base cases you can verify by hand ({W}_{1}=1, {W}_{2}=2) and count forward.

Alternative check by cases: with k two-steps and 10-2k one-steps, the count is \sum_{k=0}^{5}{\left( \genfrac{}{}{0pt}{}{10-k}{k}\right)}=1+9+28+35+15+1=89 ✔

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